All categories

2 years ago

What is the displacement of the runner in 4 laps? * 2 point

Physics

1 answer:

Serhud [2]2 years ago

3 0

Answer:

There is no displacement.

Explanation:

Because the runner is running laps and returning to the original place, there is no displacement as displacement is relative to the change in location from the original position.

Hope this helps. . .

ly UwU

You might be interested in

How do you give an example for finding net force?

Crazy boy [7]

-- 6 people all trying to push a car out of snow

-- a Tug-o-War with 30 people of different sizes pulling on each end of the rope

-- you and your sister both pulling on the same doll (or Transformer)

-- lifting a book up from the table to a high shelf
taking a book down from a high shelf to the table
(one force is you; another force is gravity)

-- grabbing your big dog by his collar and trying to bring him inside

-- three people at the table all grab the ketchup bottle at the same time

7 0

3 years ago

You kick a soccer ball straight up into the air with a speed of 21.2 m/s. How long does it take the soccer ball to reachbits hig

Dimas [21]

Gravity increased the downward speed (or decreases the upward speed) by 9.8 m/s every second.

21.2/9.8 = 2.2 seconds

7 0

3 years ago

Read 2 more answers

A small mailbag is released from a helicopter that is descending steadily at 2.52 m/s. (a) After 4.00 s, what is the speed of th

gogolik [260]

Answer: 41.72m/s

Explanation: v= u + gt

V = 2.52 + 9.8(4.00)

V = 41.72m/s

6 0

2 years ago

A tuning fork has a frequency of 280 Hz and the wavelength of the sound produced is 1.5 meters. Calculate the wave's frequency a

s2008m [1.1K]

Answer:

The answer would be 420 m/s

Explanation:

Look in attachment ⬇

I Hope this Helps!!!

3 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago